Question

How many terms of arithmetic progression 45,39,33,must be taken so that their sum is 180

Answer 1

Hello there ...

here a = 45 , d = -6 , Sn = 180

To find : n

We know ,

$sn \: = \frac{n}{2} \times 2a + (n - 1)d$

Therefore,

180 = n / 2 × 2 ( 45) + (n - 1 ) -6

180 = n/2 × 90 - 6n + 6

180 × 2 = n ( 96 - 6n )

360 = 96n - 6n^2

6n^2 - 96n + 360 = 0

Dividing by 6

n^2 - 16n + 60 = 0

n^2 - 10n - 6n + 60 = 0 .......( factorizing )

n ( n - 10 ) - 6 ( n - 10 ) = 0 .......{ taking common out }

( n - 6 ) (n - 10 ) = 0

n - 6 = 0 or. n - 10 = 0

n = 6. or. n = 10

Therefore 6 or 10 terms must be taken so that their sum is 180
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Thank You !